python etree with xpath and namespaces with prefix

Question

I can't find info, how to parse my XML with namespace:

I have this xml:

<par:Request xmlns:par="http://somewhere.net/actual">
  <par:actual>blabla</par:actual>
  <par:documentType>string</par:documentType>
</par:Request>

And tried to parse it:

dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()
for subtag in rootxml.xpath(u'//par:actual'):
    #do something
    print(subtag)

And got exception, because it doesn't know about namespace prefix. Is there best way to solve that problem, counting that script will not know about file it going to parse and tag is going to search for?

Searching web and stackoverflow I found, that if I will add there:

namespace = {u'par': u"http://somewhere.net/actual"}
for subtag in rootxml.xpath(u'//par:actual', namespaces=namespace):
    #do something
    print(subtag)

That works. Perfect. But I don't know which XML I will parse, and searching tag (such as //par:actual) is also unknown to my script. So, I need to find way to extract namespace from XML somehow.

I found a lot of ways, how to extract namespace URI, such as:

print(rootxml.tag)
print(rootxml.xpath('namespace-uri(.)'))
print(rootxml.xpath('namespace-uri(/*)'))

But how should I extract prefix to create dictionary which ElementTree wants from me? I don't want to use regular expression monster over xml body to extract prefix, I believe there have to exist supported way for that, isn't it?

And maybe there have to exist some methods for me to extract by ETree namespace from XML as dictionary (as ETree wants!) without hands manipulation?

Answer 1

You cannot rely on the namespace declarations on the root element: there is no guarantee that the declarations will even be there, or that the document will have the same prefix for the same namespace throughout. Assuming you are going to have some way of passing the tag you want to search (because you say it is not known by your script), you should also provide a way to pass a namespace mapping as well. Or use the James Clark notation, like {http://somewhere.net/actual}actual (the ETXPath has support for this syntax, whereas "normal" xpath does not, but you can also use other methods like .findall() if you don't need full xpath)

If you don't care for the prefix at all, you could also use the local-name() function in xpath, eg. //*[local-name()="actual"] (but you won't be "really" sure it's the right "actual")

Answer 2

Oh, I found it.

After we do that:

dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()

Object rootxml contains dictionary nsmap, which contains all namespaces that I want.

So, simplest solution I've found:

dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()
nss = rootxml.nsmap
for subtag in rootxml.xpath(u'//par:actual', namespaces=nss):
    #do something
    print(subtag)

That works.

UPD: that works if user understand what means 'par' in XML he works with. For example, comparing supposed namespace with existing namespace before any other operations.

Still, I like much variant with XPath that understands {...}actual, that was what I tried to achieve.